martingale math - what say ye?

So I get that martingale system is mathematically sound, but in real world it doesn't work because casinos set table limits, etc.

But out of curiosity for you mathematical minded folk, at what point would it work?

For example, it doesn't work in roulette because most tables have low maximums, and the chances of black hitting (for example) in a large series (although unlikely statistically speaking) is going to happen frequently.

That is in a system where probability of success/failure is 50/50 (or very close to it).

So the question is, what ratio would you say that system could possibly work practically speaking? For example, lets say the probability of success of single trial is 2/3, and the chance of of loosing is 1/3. Would you think you could use martingale in that setting? If not, how about .75 chance of probability of success, vs .25 of losing?


Even more real world, starting with $7000, if you lay a $20 bet on 4 (0.083333333% chance of hitting a 4 hitting each roll), a 4 would need to hit 6 times before losing your bank role (20+40+120+360+1080+3240). Using the binomial formula, the chances of a 4 hitting 6 times lets say in 12 roles is 0.000198744170760046. However, the chance of a 7 coming up in 12 roles is 0.887843345215385.

That seems pretty good. However, I could never lay down that much money (don't have the balls), and I also know that casino's know math better than I do.

What if I could lay a 2, would the martingale system work then? At what odds in your favor would real mathematicians take in order to say they had a fighting chance with the martingale?

Quote: slackyhacky

So I get that martingale system is mathematically sound, but in real world it doesn't work because casinos set table limits, etc.

No, it's not mathematically sound. It's only neutral (not advantageous) for a fair bet.

In casino games, martingale increases the effective house edge. I'll quote one of my older posts on this.

...
For n-deep martingale in roulette, the possible outcomes of a martingale progression are:
1-(20/38)^n: Win 1 unit
(20/38)^n: Lose 2^n-1 units
For a total EV of 1-(20/38)^n-(2^n-1)(20/38)^n = 1-(20/38)^n*2^n = 1-(40/38)^n ~= 1-1.0526^n

Or, for other games, 1-(1+HA)^n.
It is a very simple function. For any HA>0, it exponentially approaches -infinity as possible martingale progression depth (n) increases.
If you just martingale 2 bets deep, you are playing a 10.8% house edge game instead of 5.26%. If you do it 3 bets deep, 16.7%. Playing $10-$1,000 game will let you martingale 7 bets deep, for a house edge of 43%.

In other words, you expect to encounter almost twice as quickly a row of 7 losing bets opposed to a fair 50/50 game, because of that seemingly "little" house advantage....

Quote: P90

No, it's not mathematically sound. It's only neutral

Ditto.

To be able to have even a small shot at it "working", you'd need both an unlimited bankroll, and a table with no maximum bet.

It doesn't matter what the game is.

Quote: DJTeddyBear

To be able to have even a small shot at it "working", you'd need both an unlimited bankroll, and a table with no maximum bet.

Which of course is a situation to be found only in mathematical textbooks and debating societies. In the real world this "Double Up to Catch Up" would always encounter rather limited bankrolls and would most definitely encounter table limits.

Sure, some "short run" statistics can be generated wherein it "works" adequately but in the real world there is always that "nth" spin wherein red appears in succession simply too many times for the system to work.

If you have ANY advantage, as the OP is asking, then you should Kelly bet, not Martingale. Kelly bet grinds out the advantage using bets proportional to the size of your bankroll.

Martingaling an advantage will not win as much in the long term as Kelly betting.

The leftover question is "under what circumstances would Martingaling still have an advantage?" I don't care to find out.

Quote: dwheatley

The leftover question is "under what circumstances would Martingaling still have an advantage?"

Doubling your bet after a loss in any succession of losses would work only under the circumstances of your knowing when the losses would end and if you knew that, you wouldn't have any losses to begin with.

Quote: P90

No, it's not mathematically sound. It's only neutral (not advantageous) for a fair bet.

How is that true. If A or B are the outcomes, and everytime B happens, it clears the debt. Everytime B happens consequetively, I gain. Over time, that gain persists.

Quote: P90

...
For n-deep martingale in roulette, the possible outcomes of a martingale progression are:
1-(20/38)^n: Win 1 unit
(20/38)^n: Lose 2^n-1 units
For a total EV of 1-(20/38)^n-(2^n-1)(20/38)^n = 1-(20/38)^n*2^n = 1-(40/38)^n ~= 1-1.0526^n

Or, for other games, 1-(1+HA)^n.
It is a very simple function. For any HA>0, it exponentially approaches -infinity as possible martingale progression depth (n) increases.
If you just martingale 2 bets deep, you are playing a 10.8% house edge game instead of 5.26%. If you do it 3 bets deep, 16.7%. Playing $10-$1,000 game will let you martingale 7 bets deep, for a house edge of 43%.

I don't follow this at all. First of all, where are you getting a house edge of 5.26%? I get lost in the math quickly so you tell me if this is wrong. Assuming a 5% vig on 20 dollars, I bet $21 to win $10. 10/21= .47619. The odds of a 7 coming up before a 10 is exactly 2/1 (or 0.5). So the house advantage is about 2.38%.

Also, using my example, after lossing the series I mentioned (20+40+120+360+1080+3240) - the 5% vig on each of those comes out to be 243. How are you coming up with 43%?

So the question was, in what odds (and when I say odds, I mean in your favor for probability of the event happening, not on the house edge, because you can easily add that into your losses since you know as sure as the sun comes up that your event is coming that will clear debt) in your favor would you feel comfortable playing the martingale system, and the resounding answer is NONE?


Interesting......

Quote: slackyhacky

How is that true. If A or B are the outcomes, and everytime B happens, it clears the debt.

B doesn't have to happen in your play session at all, even once. Since your bankroll is limiting it.

Quote: slackyhacky

I don't follow this at all. First of all, where are you getting a house edge of 5.26%?

It's for roulette. For craps, use whatever it is.